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#### Albert

##### Well-known member

- Jan 25, 2013

- 1,225

$c+1=2a^2$

$c^2+1=2b^2$

c is a prime

a,b,c are roots of $ 3x^3+rx^2+sx+t=0 $

please find r and t

- Thread starter Albert
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- Thread starter
- #1

- Jan 25, 2013

- 1,225

$c+1=2a^2$

$c^2+1=2b^2$

c is a prime

a,b,c are roots of $ 3x^3+rx^2+sx+t=0 $

please find r and t

- May 31, 2013

- 118

It may not be an ellegent one i just gave it a shot$$

\begin{align*}c^2+2c+1&=4a^4\\c&=2a^4-b^2\end{align*}$$

I gave a try at a=2 so b is 5 and c is 7.

And 7 is also the first prime to satisfy both equations even individually too.

So r is -52 and t is -210

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- #3

- Feb 7, 2012

- 2,725

$c+1=2a^2$

$c^2+1=2b^2$

c is a prime

a,b,c are roots of $ 3x^3+rx^2+sx+t=0 $

please find r and t

I think that it is, but I can't prove it. In fact, the equation $c^2+1=2b^2$ can be written $\bigl(\frac cb\bigr)^2 + \bigl(\frac 1b\bigr)^2 = 2$, so that $c/b$ is a lower convergent for $\sqrt2$. The condition $c+1=2a^2$ requires that $\frac12(c+1)$ is a square, and the only Pell number less than $10^9$ satisfying that condition is $7$. So any other solution for $c$ must be bigger than that.

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- #4

- Jan 25, 2013

- 1,225

$a,b,c\in N,\,\, c:a\,\, prime$

$c+1=2a^2---(1)$

$c^2+1=2b^2---(2)$

(2)-(1):$c(c-1)=2(b-a)(b+a)---(3)$

from (1)(2)(3) we conclude :

$c\geq b \geq a,\,\, and \,\,\,\, c \mid (a+b),( or\,\, (a+b) \,\, mod \,\, c=0)--(4)$

$\therefore c=a+b, \rightarrow b=c-a$

from (3)

$c-1=2b-2a=2(c-a)-2a=2c-4a$

$\therefore c+1=4a=2a^2$

$\therefore a=2,c=7,b=5$

$\therefore r=-3\times (2+5+7)=-42,\,\, t=-3\times(2\times5\times7)=-210$

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